JEE Main & Advanced Chemistry Electrochemistry / विद्युत् रसायन Question Bank Electrode potential Ecell Nernst equation and ECS

  • question_answer
    Adding powdered lead and iron to a solution that is 1.0 M in both \[P{{b}^{2+}}\]and \[F{{e}^{2+}}\]ions, would result a reaction, in which [CPMT 1987]

    A)                 More iron and \[P{{b}^{2+}}\]ions are formed

    B)                 More lead and \[F{{e}^{2+}}\]ions are formed

    C)                 Concentration of both \[P{{b}^{2+}}\]and \[F{{e}^{2+}}\]ions increases

    D)                 There is no net change

    Correct Answer: D

    Solution :

               \[\Delta {{G}^{o}}=-n{{E}^{o}}F\]            \[F{{e}^{2+}}+2{{e}^{-}}\to Fe\]                                    ?..(i)            \[\Delta {{G}^{o}}=-2\times F\times (-0.440\,V)=0.880\,F\]            \[F{{e}^{3+}}+3{{e}^{-}}\to Fe\]                                    ?..(ii)            \[\Delta {{G}^{o}}=-3\times F\times (-0.036)=0.108F\]            On subtracting equation (i) from (ii)            \[F{{e}^{3+}}+{{e}^{-}}\to F{{e}^{2+}}\]            \[\Delta {{G}^{o}}=0.108F-0.880F=-0.772F\]                 \[{{E}^{o}}\] for the reaction \[=-\frac{\Delta {{G}^{o}}}{nF}\]    \[=-\frac{(-\,0.772\,F)}{1\times F}=+\,0.772\,V\].


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